Maximum Nodes at Any Level in Binary Tree

Unit 9 Session 2 (Click for link to problem statements)

Problem Highlights

💡 Difficulty: Easy
⏰ Time to complete: 20+ mins
🛠️ Topics: Tree, Breadth-First Search, Queue, Level Order Traversal

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

Established a set (2-3) of test cases to verify their own solution later.

Established a set (1-2) of edge cases to verify their solution handles complexities.

Have fully understood the problem and have no clarifying questions.

Have you verified any Time/Space Constraints for this problem?

Can the tree be empty?
- Yes, if the tree is empty, return 0.
What should be returned if the tree has only one node?
- Return 1, as there is only one node at the first level.

HAPPY CASE
 
      5 (root)
     / \
    3   6
   / \   \
  2   4   7
 /
1   

Input: root   
Output: 3
Explanation: The third level has the most nodes (3 nodes).

EDGE CASE
Input: None
Output: 0
Explanation: The tree is empty, so return 0.

2: M-atch

Match what this problem looks like to known categories of problems, e.g., Linked List or Dynamic Programming, and strategies or patterns in those categories.

For Tree problems, we want to consider the following approaches:

Breadth-First Search (BFS): Useful for level order traversal.
Queue: Used to manage the order of node exploration.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use a BFS approach to traverse the tree level by level. Use a queue to keep track of nodes to be explored, and an integer variable to store the maximum number of nodes at any level.

1) If the tree is empty, return 0.
2) Create an empty queue and add the root node.
3) Initialize a variable to store the maximum number of nodes at any level to 0.
4) While the queue is not empty:
    a) Get the number of nodes at the current level.
    b) Update the maximum number of nodes if the current level size is greater.
    c) For each node at the current level:
        i) Pop the node from the queue.
        ii) Add the left child to the queue if it exists.
        iii) Add the right child to the queue if it exists.
5) Return the maximum number of nodes at any level.

⚠️ Common Mistakes

Forgetting to handle the case where the tree is empty.
Not correctly managing the queue for BFS.
Incorrectly updating the maximum number of nodes for each level.

4: I-mplement

Implement the code to solve the algorithm.

from collections import deque

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def level_max(root):
    if not root:
        return 0
    
    max_nodes = 0
    queue = deque([root])
    
    while queue:
        level_size = len(queue)
        max_nodes = max(max_nodes, level_size)
        
        for _ in range(level_size):
            node = queue.popleft()
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
    
    return max_nodes

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

Trace through your code with an input to check for the expected output
Catch possible edge cases and off-by-one errors
For a starting point, try checking your code against the Happy/Edge Case(s) above

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Time Complexity:

O(N), where N is the number of nodes in the tree.
This is because we must visit each node exactly once to count the nodes at each level.

Space Complexity:

O(N), where N is the number of nodes in the tree.
This is because the queue could hold up to N nodes in the worst case (when the tree is completely unbalanced).

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